A metallic bar is heated from 0° C to 100°C . The coeficient of linear expansion is 10-5 K-1. What will be the percentage increase in length?

Question

A metallic bar is heated from 0° C to 100°C . The coeficient of linear expansion is 10-5 K-1. What will be the percentage increase in length? (A) 0.01 % (B) 0.1 % (C) 1 % (D) 10 %

Tardigrade Admin · Accepted Answer

(Δ l/l) = α Δ T = 10-5 ×100 = 10-3 (Δ l/l) ×100 % = 10-3 ×100 = 10-1 = 0.1 %

Q. A metallic bar is heated from $0^{\circ} C$ to $10 0^{\circ} C$ . The coeficient of linear expansion is $1 0^{- 5} K^{- 1}$ . What will be the percentage increase in length?

0.01 %

0.1 %

1 %

10 %

$\frac{Δ l}{l} = α Δ T = 1 0^{- 5} \times 100 = 1 0^{- 3}$
$\frac{Δ l}{l} \times 100% = 1 0^{- 3} \times 100$
$= 1 0^{- 1} = 0.1%$

Q. A metallic bar is heated from 0∘C to 100∘C. The coeficient of linear expansion is 10−5K−1. What will be the percentage increase in length?

0.01 %

0.1 %

1 %

10 %

lΔl​=αΔT=10−5×100=10−3 lΔl​×100%=10−3×100 =10−1=0.1%

Q. A metallic bar is heated from $0^{\circ} C$ to $10 0^{\circ} C$ . The coeficient of linear expansion is $1 0^{- 5} K^{- 1}$ . What will be the percentage increase in length?

$\frac{Δ l}{l} = α Δ T = 1 0^{- 5} \times 100 = 1 0^{- 3}$
$\frac{Δ l}{l} \times 100% = 1 0^{- 3} \times 100$
$= 1 0^{- 1} = 0.1%$